A metal worker wants to heat commercial bronze (a copper alloy) by using an air cannon pointed downward. If she wants to increase the temperature of the bronze by 1.5 o F and the cannon is 20 feet above the ground, what would be exit velocity from the cannon need to be (in ft/s)? Neglect air resistance. [V = 78.9 ft/s]

exit velocity from the cannon is 78.9 ft/s

Explanation:

Given that;

increase the temperature of the bronze by 1.5 o F

cannon is 20 feet above the ground.

we know that

( Kinetic energy + potential energy )at state1 = Internal energy

1/2mv² + mgh = mCpΔT

V²/2 + gh = CpΔT

V²/2 = CpΔT - gh

V² = 2(CpΔT - gh)

V = √(2(CpΔT - gh))

So we substitute

V = √(2((0.1 × 778.16 × 32.174 × 1.5) - (32.17 × 20)))

V = √(2(3755.4779 - 643.4))

V = √(2 × 3112.0779)

V = √6224.1558

V = 78.89 ≈ 78.9

Therefore exit velocity from the cannon is 78.9 ft/s

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